chapter 5 motion
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Chapter 5 MotionChapter 5 Motion
Average VelocityTimeTotal
DistanceTotal
if
if
tt
dd
t
d
V
if
if
tt
vv
t
v
a
if t1 = 0,
t
vv if a
d = v t
Chapter 5 MotionChapter 5 Motion
t
vv if a Solve for vf
atvv if
Chapter 5 MotionChapter 5 Motion
tvv
d fi
2
but 2
fi vvv
tvd
Chapter 5 MotionChapter 5 Motion
tvv
d fi
2but atvv if
tv
d i
2
atvi tatv
d i
2
2
2
2
1attvd i
Chapter 5 MotionChapter 5 Motion
tvv
d fi
2atvv if
Solve both equations for t
fi vv
dt
2a
vvt if
Chapter 5 MotionChapter 5 Motion
fi vv
dt
2
a
vvt if
Set equal to each other
fi vv
d2
a
vv if
Chapter 5 MotionChapter 5 Motion
fi vv
d2
a
vv if Multiply across
222 if vvad
advv if 222
Chapter 5 MotionChapter 5 Motion
advv if 222
2
2
1attvd i
d = v t
t
vv if a
tvv
d fi
2
atvv if
Chapter 5 MotionChapter 5 Motion
A car is traveling at 65 km/hr. The brakesare applied and the car stops in 3 seconds.a. What is the car’s acceleration?b. How far does it travel?
vi = 65 km/hrt = 3 svf = 0
vi = 65 km/hr = 18m/s
Chapter 5 MotionChapter 5 Motion
A car is traveling at 65 km/hr. The brakesare applied and the car stops in 3 seconds.a. What is the car’s acceleration?b. How far does it travel?
vi = 18 m/st = 3 svf = 0s
sm
t
vva if
3
/180
2/6 sma
Chapter 5 MotionChapter 5 Motion
A car is traveling at 65 km/hr. The brakesare applied and the car stops in 3 seconds.a. What is the car’s acceleration?b. How far does it travel?
2
2
1attvd i
22 )3)(/6(2
1)3)(/18( ssmssmd =27 m
Chapter 5 MotionChapter 5 Motion
An airplane must reach a speed of 55 m/sbefore take off. It can accelerate at 12 m/s2.a. How long does the runway have to be?b. How long will this take?
vi = 0a = 12 m/s2
vf = 55 m/s
advv if 222
ma
vvd if 126
2
22
Chapter 5 MotionChapter 5 Motion
An airplane much reach a speed of 55 m/sbefore take off. It can accelerate at 12 m/s2.a. How long does the runway have to be?b. How long will this take?
vi = 0a = 12 m/s2
vf = 55 m/sa
vv if t
ssm
smsm58.4
/12
/0/552
t
Chapter 5 MotionChapter 5 Motion
Scott is driving a car at 20 m/s and notices a cow in front of him at a distance of 80 m away. He brakes at a rate of 3 m/s2. What happens to the cow?
vi = 20 m/sa = -3 m/s2
vf = 0
Chapter 5 MotionChapter 5 Motion
a
vvd if
2
22
)/3(2
)/20(02
2
sm
smd
md 7.66
The cowsurvives!!!!
Acceleration Due to Gravity
ag = -9.8 m/s2
For all free falling bodies on earth, theyacceleratedownward at 9.8 m/s2.
Acceleration Due to Gravity
A ball is thrown upward at 49 m/s.a. How long is it in the
air?b. How high does it
go?c. How fast is it going
when it lands?
Acceleration Due to Gravity
How long is it in the air? vi = 49 m/sa = -9.8 m/s2
vf = 0a
vv if t 2/8.9
/490
sm
sm
t
sec5t Total time in air is 10 sec
Acceleration Due to Gravity
How high does it go?
vi = 49 m/sa = -9.8 m/s2
t = 5 sec
2
2
1attvd i
22 )5)(/8.9(2
1)5)(/49( ssmssmd
md 5.122
Acceleration Due to Gravity
How fast is it going when it lands?
vi = 49 m/sa = -9.8 m/s2
t = 10 secatvv if
)10)(/8.9(/49 2 ssmsmv f
smv f /49
A skier travels from A to B to C to D.What is her average velocity and speed?
When a traffic light turns green, a waiting car starts off with a constant acceleration of 6 m/s2. At the instant the car begins to accelerate, a truck with constant velocity of 21 m/s passes in the next lane.
a. How far will the car travel before it overtakes the truck?
b. How fast will the car be traveling when it overtakes the truck?
c. Construct a distance vs time graph for the car and the truck on the same graph.
Stoplight Problem
a. How far will the car travel before it overtakes the truck?
• Truck d = vt Car d = vit + ½ at2
• d = 21 t d = ½ (6) t2
• 21t = 3t2
• 3t2 – 21t = 0• 3t(t – 7) = 0 t = 0 s, 7 s
Stoplight Problem
b. How fast will the car be traveling when it overtakes the truck?
•d = 21 t = 21(7) = 147m •d = ½ at2 = (1/2)(6)(7)2 = 147m
c. Construct a distance vs time graph for the car and the truck on the same graph.
• Graph y1 = 21t, y2 = 3t2 with the –0 symbol. Under mode select simultaneous with a window of [0,10],[0,150]
Stoplight Problem
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